Clayton College & State University
Department of Natural Sciences
September 8, 2004
Physics 1112 – Quiz 2
Name ____SOLUTION_________________________________
Three charged particles are arranged on the line as shown below. Calculate the total
electric force on particle 3 (the – 4.00 C on the right) due to the other two charges.
F13= ke|Q1||Q3 | / r12 = (8.99 x 109 Nm2/C2)(8.00 x 10–6C)(4.00 x 10–6C)/(0.500 m)2 = 1.15
N to the right
F23= ke|Q2||Q3 | / r12 = (8.99 x 109 Nm2/C2)(3.00 x 10–6C)(4.00 x 10–6C)/(0.200 m)2 = 2.70
N to the left
F3= F13+ F23= 1.15 N -2.70 N = -1.55 N
Clayton State University
Department of Natural Sciences
August 31, 2007
Physics 1112 – Quiz 2
Name ___SOLUTION__________________________________
1.
Particles of charge 50.0 mC, - 60.0 mC and 70.0 mC are placed in a line as shown
below. The center one is 0.500 m from each of the others. Calculate the net force
acting on the central charge due to the other two.
q1 = +50.0 x 10-3 C
q2 = -60.0 x 10-3 C
q3 = +70.0 x 10-3 C
F12 = k | q1|| q2|/r122 = (8.99 x 109 N-m2/C2) (50.0 x 10-3 C) (60.0 x 10-3 C) / (0.500 m)2 = 1.08 x
108 N, to the left
F32 = k |q2 ||q3|/r232 = (8.99 x 109 N-m2/C2) (60.0 x 10-6 C) (70.0 x 10-6 C) / (0.500 m)2 = 1.51 x
108 N, to the right
Ftot = F12 + F32= - 1.08 x 108 N + 1.51 x 108 N = 4.30 x 107 N (to the right)
Department of Natural Sciences
Clayton State University
Department of Natural Sciences
January 26, 2006
Physics 1112 – Quiz 2
Name ___SOLUTION__________________________________
A charge of 6.00 mC is placed at each corner of a square 0.100 m on a side.
Determine the magnitude and direction of the force on charge q2.
F12 = k q1q2/r122 = (8.99 x 109 N-m2/C2) (6.00 x 10-3 C) (6.00 x 10-3 C) / (0.100 m)2 = 3.24 x 107
N, to the right
F32 = k q3q2/r322 = (8.99 x 109 N-m2/C2) (6.00 x 10-3 C) (6.00 x 10-3 C) / (0.100 m)2 = 3.24 x 107
N, upward
F42 = k q4q2/r422 = (8.99 x 109 N-m2/C2) (6.00 x 10-3 C) (6.00 x 10-3 C) / (0.141 m)2 = 1.62 x 107
N, at 45o with the positive x-axis
F12x = F12 cos(0) = 3.24 x 107 N
F12y = F12 sin(0) = 0
F32x = F32 cos(90o) = 0
F32y = F32 sin(90o) = 3.24 x 107 N
F42x = F42 cos(180o) = 1.14 x 107 N
F42y = F42 sin(180o) = 1.14 x 107 N
Ftotx = F12x + F32x + F42x = 4.38 x 107 N
Ftoty = F12y + F32y + F42y= 4.38 x 107 N
Ftot = (Ftotx2 + Ftoty2)1/2 = 6.20 x 107 N
= tan-1(Ftoty /Ftotx) = 45o
Clayton State University
Department of Natural Sciences
January 23, 2007
Physics 1112 – Quiz 2
Name ___SOLUTION__________________________________
1.
Particles of charge 75,  48, and 85 C are placed in a line as shown below. The
center one is 0.350 m from each of the others. Find the electric force on the +75 C
charge. Do not forget to specify magnitude and direction.
q1 = +75.0 x 10-6 C
q2 = +48.0 x 10-6 C
q3 = -85.0 x 10-6 C
F21 = k | q1|| q2|/r12 = (8.99 x 109 N-m2/C2) (75.0 x 10-6 C) (48.0 x 10-6 C) / (0.350 m)2 = 264 N, to
the left
F31 = k |q1 ||q3|/r22 = (8.99 x 109 N-m2/C2) (48.0 x 10-6 C) (85.0 x 10-6 C) / (0.700 m)2 = 117 N, to
the right
Ftot = F21 + F31= -264 N + 117 N = -147 N (to the left)
Clayton State University
Department of Natural Sciences
January 24, 2008
Physics 1112 – Quiz 2
Name __SOLUTION___________________________________
1.
Particles of charge 50.0 mC, - 60.0 mC and 70.0 mC are placed in a line as shown
below. The center one is 0.500 m from each of the others. Calculate the net electric
field due to the charges 0.500 m to the right from third charge.
q1 = +50.0 x 10-3 C
q2 = -60.0 x 10-3 C
q3 = +70.0 x 10-3 C
E1 = k | q1|/r12 = (8.99 x 109 N-m2/C2) (50.0 x 10-3 C)/(1.50 m)2 = 2.00 x 108 N/C, to the right
E2 = k |q2 |/r22 = (8.99 x 109 N-m2/C2) (60.0 x 10-3 C)/(1.00 m)2 = 5.39 x 108 N/C, to the left
E3 = k | q3|/r32 = (8.99 x 109 N-m2/C2) (70.0 x 10-3 C)/(0.500 m)2 = 25.2 x 108 N/C, to the right
Etot = E1 + E2 + E3 = 2.00 x 108 N/C - 5.39 x 108 N/C + 25.2 x 108 N/C = 2.18 x 109 N/C (to the
right)
Clayton State University
Department of Natural Sciences
June 7, 2006
Physics 1112 – Quiz 2
Name _SOLUTION____________________________________
1. Particles of charge 75,  48, and 85 C are placed in a line as shown below. The
center one is 0.350 m from each of the others. Calculate the net forces acting on
two end charge due to the other two.
q1 = +75.0 x 10-6 C
q2 = +48.0 x 10-6 C
q3 = -85.0 x 10-6 C
F21 = k |q2 || q1|/r212 = (8.99 x 109 N-m2/C2) (48.0 x 10-6 C) (75.0 x 10-6 C) / (0.350 m)2 = 264 N,
to the left
F31 = k |q3 || q1|/r312 = (8.99 x 109 N-m2/C2) (85.0 x 10-6 C) (75.0 x 10-6 C) / (0.700 m)2 = 117 N,
to the right
F1 = F31 + F21 = 117 N – 264 N = -147 N (to the left)
F23 = k |q2 || q3|/r232 = (8.99 x 109 N-m2/C2) (48.0 x 10-6 C) (85.0 x 10-6 C) / (0.350 m)2 = 299 N,
to the left
F13 = k |q1 || q3|/r132 = (8.99 x 109 N-m2/C2) (85.0 x 10-6 C) (75.0 x 10-6 C) / (0.700 m)2 = 117 N,
to the left
F3 = F13 + F23 = - 117 N – 299 N = - 416 N (to the left)
Clayton State University
Department of Natural Sciences
June 5, 2007
Physics 1112 – Quiz 2
Name _____SOLUTION________________________________
1.
Particles of charge – 25.0, + 47.0 and – 65.0 C are placed in a line as shown below.
The center one is 0.550 m from each of the others. Find the electric force on the
+47.0 C charge. Do not forget to specify magnitude and direction.
q1 = -25.0 x 10-6 C
q2 = +47.0 x 10-6 C
q3 = -65.0 x 10-6 C
F12 = k | q1|| q2|/r122 = (8.99 x 109 N-m2/C2) (25.0 x 10-6 C) (47.0 x 10-6 C) / (0.550 m)2 = 34.9 N,
to the left
F32 = k |q2 ||q3|/r232 = (8.99 x 109 N-m2/C2) (47.0 x 10-6 C) (65.0 x 10-6 C) / (0.550 m)2 = 90.8 N,
to the right
Clayton State University
Department of Natural Sciences
June 3, 2008
Physics 1112 – Quiz 2
Name _____SOLUTION________________________________
1.
Particles of charge -20.0 C, +35.0 Cand - 56.0 C are placed in a line as shown
below. The center one is 0.500 m from each of the others. Calculate the net electric
field due to the charges 0.500 m to the right from third charge.
q1 = -20.0 x 10-6 C
q2 = +35.0 x 10-6 C
q3 = -56.0 x 10-6 C
E1 = k | q1|/r12 = (8.99 x 109 N-m2/C2) (20.0 x 10-6 C)/(1.50 m)2 = 0.799 x 105 N/C, to the left
E2 = k |q2 |/r22 = (8.99 x 109 N-m2/C2) (35.0 x 10-6 C)/(1.00 m)2 = 3.14 x 105 N/C, to the right
E3 = k | q3|/r32 = (8.99 x 109 N-m2/C2) (56.0 x 10-6 C)/(0.500 m)2 = 20.1 x 105 N/C, to the left
Etot = E1 + E2 + E3 = -0.799 x 105 N/C + 3.14 x 105 N/C - 20.1 x 105 N/C = -1.78 x 106 N/C (Net
electric field points to the left.)